Show that 1−cos2x1+cos2x≡tan2x\displaystyle \frac{1 - \cos 2x}{1 + \cos 2x} \equiv \tan^2 x1+cos2x1−cos2x≡tan2x
Hence solve, for −2π≤θ≤2π-2\pi \leq \theta \leq 2\pi−2π≤θ≤2π,
1−cos2θ1+cos2θ=3 \frac{1 - \cos 2\theta}{1 + \cos 2\theta} = 3 1+cos2θ1−cos2θ=3348 exam-style questions on Edexcel A Level Maths Trigonometry and Modelling, covering 7.1 Addition Formulae, 7.2 Double Angle Formulae, 7.3 Solving Trigonometric Equations, 7.4 Simplifying a cos x +- b sin x, 7.5 Proving Trigonometric Identities, and 7.6 Modelling with Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.